Finding the splitting field of $x^6-8$ over $\mathbb{Q}$ Finding the splitting field of $x^6-8$ over $\mathbb{Q}$.
So, this polynomial factors as $(x^2-2)(x^4+2x^2+4)$ and so all the roots will be $\pm \sqrt{2}$ and $\pm \sqrt{-1 \pm i\sqrt{3}}$. So a splitting field would be $\mathbb{Q}(\sqrt{-1 \pm i\sqrt{3}},\sqrt{2})$.
How do I tell if these added roots are linearly independent? Or if adding $\sqrt{-1 + i\sqrt{3}}$ includes $\sqrt{-1 - i\sqrt{3}}$? And then once I get the splitting field written in it's nicest form can somebody help me calculate the degree of the extension? I'd really appreciate it thanks!
 A: Let $a=\sqrt{2}$, and let $b=\sqrt{-3}$.

Let $\omega=\exp\left(i\left({\large{\frac{\pi}{3}}}\right)\right)$.

Noting that


*

*$\omega$ is a primitive $6$-th root of $1$.$\\[4pt]$

*$\sqrt[6]{8}=\sqrt{2}=a$.


it follows that the roots of $x^6-8$ in $\mathbb{C}$ are
$$a,a\omega,a\omega^2,a\omega^3,a\omega^4,a\omega^5$$
hence $K=\mathbb{Q}(a,\omega)$ is a splitting field of $f$.

Then since
$$
\omega
=
\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)
=
\frac{1}{2}+\frac{1}{2}i\sqrt{3}
=
\frac{1}{2}+\frac{1}{2}b
$$
it follows that we can also write $K=\mathbb{Q}(a,b)$, or explicitly, $K=\mathbb{Q}\left(\sqrt{2},\sqrt{-3}\right)$.

Note that $[\mathbb{Q}(a):\mathbb{Q}]=2$ and $[\mathbb{Q}(b):\mathbb{Q}]=2$.

Since $a\in\mathbb{R}$, and $b\not\in\mathbb{R}$, it follows that $[\mathbb{Q}(a,b):\mathbb{Q}(a)] > 1$.

From $[\mathbb{Q}(b):\mathbb{Q}]=2$, we get $[\mathbb{Q}(a,b):\mathbb{Q}(a)]\le 2$, hence $[\mathbb{Q}(a,b):\mathbb{Q}(a)]=2$

Hence, we get
$$
[K:\mathbb{Q}]
=
[\mathbb{Q}(a,b):\mathbb{Q}]
=
[\mathbb{Q}(a,b):\mathbb{Q(a)}]
{\,\cdot\,}
[\mathbb{Q}(a):\mathbb{Q}]
=
2{\,\cdot\,}2
=
4
$$
